Python 使用 seaborn 对数记录 lmplot
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Log-log lmplot with seaborn
提问by sjdh
Can the function lmplotfrom Seaborn plot on a log-log scale?
This is lmplot on a normal scale
lmplotSeaborn 中的函数可以在对数尺度上绘制吗？这是正常规模的 lmplot
import numpy as np
import pandas as pd
import seaborn as sns
x = 10**arange(1, 10)
y = 10** arange(1,10)*2
df1 = pd.DataFrame( data=y, index=x )
df2 = pd.DataFrame(data = {'x': x, 'y': y})
sns.lmplot('x', 'y', df2)
采纳答案by mwaskom
If you just want to plot a simple regression, it will be easier to use seaborn.regplot. This seems to work (although I'm not sure where the y axis minor grid goes)
如果您只想绘制一个简单的回归，使用seaborn.regplot. 这似乎有效（虽然我不确定 y 轴小网格在哪里）
import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
x = 10 ** np.arange(1, 10)
y = x * 2
data = pd.DataFrame(data={'x': x, 'y': y})
f, ax = plt.subplots(figsize=(7, 7))
ax.set(xscale="log", yscale="log")
sns.regplot("x", "y", data, ax=ax, scatter_kws={"s": 100})
If you need to use lmplotfor other purposes, this is what comes to mind, but I'm not sure what's happening with the x axis ticks. If someone has ideas and it's a bug in seaborn, I'm happy to fix it:
如果您需要lmplot用于其他目的，这就是我想到的，但我不确定 x 轴刻度发生了什么。如果有人有想法并且这是 seaborn 中的错误，我很乐意修复它：
grid = sns.lmplot('x', 'y', data, size=7, truncate=True, scatter_kws={"s": 100})
grid.set(xscale="log", yscale="log")
回答by Paul H
Call the seaborn function first. It returns a FacetGridobject which has an axesattribute (a 2-d numpy array of matplotlib Axes). Grab the Axesobject and pass that to the call to df1.plot.
首先调用seaborn函数。它返回一个FacetGrid具有axes属性的对象（matplotlib 的二维 numpy 数组Axes）。抓取Axes对象并将其传递给对 的调用df1.plot。
import numpy as np
import pandas as pd
import seaborn as sns
x = 10**np.arange(1, 10)
y = 10**np.arange(1,10)*2
df1 = pd.DataFrame(data=y, index=x)
df2 = pd.DataFrame(data = {'x': x, 'y': y})
fgrid = sns.lmplot('x', 'y', df2)
ax = fgrid.axes[0][0]
df1.plot(ax=ax)
ax.set_xscale('log')
ax.set_yscale('log')
